Chiều mk lm cho

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Ta có:

\(\sqrt{x^2-2018x+2018}+\sqrt{x^2-1009x+1009}=2x\)

\(\Leftrightarrow x-\sqrt{\left(2018x-2018\right)}+x-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow2x-\sqrt{\left(2018x-2018\right)}-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow\sqrt{\left(2018x\right)-2018}+\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow\sqrt{\left(2018x-2018\right)}=\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow2018x-2018=1009x-1009=0\Leftrightarrow x=1\)

tau méc thầy hùng

\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)

\(=\frac{11}{16}:\frac{11}{32}\)

\(=\frac{11}{16}.\frac{32}{11}\)

\(=2\)

Ta có a^2018 + b^2018 +c^2108 = a^1009b^1009 + b^1009c^1009 +c^1009a^1009

       => a^2018 + b^2018 +c^2018 -a^1009b^1009 -b^1009c^1009 -c^1009a^1009 =0

      => 2( a^2018 +b^2108 +c^2018 -a^1009b^1009 -b^1009c^1009 -c^1009a^1009) =0

      => [(a^1009)^2 -2a^1009b^1009 +(b^1009)^2] + [(b^1009)^2 -2b^1009c^1009 +(c^1009)^2] +[(c^1009)^2 -2c^1009a^1009 +(c^1009)^2] =0

     => (a^1009 -b^1009)^2 + (b^1009 -c^1009)^2 + (c^1009 -a^1009)^2 =0

   Vì (a^1009 -b^1009)^2 , (b^1009-c^1009)^2 , (c^1009- a^1009)^2 >_0 ( với mọi a,b,c)

    => a^1009 -b^1009 =0 , b^1009-c^1009 =0 , c^1009-a^1009 =0

   => a=b=c=0

 Thay vào A : A=0

Vậy A=0

2 x 2017 mũ 1009 lớn hơn vì 2017 mũ 1009 sẽ được thêm gấp đôi => sẽ lớn hơn

Ta có: \(x^2+y^2=1\Leftrightarrow\left(x^2+y^2\right)^2=1\)  (1)

Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được:

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right)ab\)

\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)

\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)

\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)

\(\Leftrightarrow x^2b-y^2a=0\)

\(\Leftrightarrow x^2b=y^2a\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1009}=\left(\frac{y^2}{b}\right)^{1009}=\left(\frac{1}{a+b}\right)^{1009}\)

\(\Rightarrow\frac{x^{2018}}{a^{1009}}=\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}\)

\(\Rightarrow\frac{x^{2018}}{a^{1009}}+\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}+\frac{1}{\left(a+b\right)^{1009}}=\frac{2}{\left(a+b\right)^{1009}}\left(đpcm\right)\)

đè bài của t sái

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{\left(a+b\right)}\) Dề ntn thế này mới chuẩn >:

Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath tham khảo

Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)

Dấu = xảy ra \(\Leftrightarrow a=b=c\)

Mà đẳng thức trên xảy ra dấu =

\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)

Bài kia tí nghĩ nốt, khó v

Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)

Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)